求值：1+cos20°2sin20°−sin10°(1tan5°−tan5°)．

Question

求值：(1+cos20°)/(2sin20°)−sin10°((1)/(tan5°)−tan5°)．

Answer 1

原式=(2cos210°)/(4sin10°cos10°)−sin10° ((cos5°)/(sin5°)−(sin5°)/(cos5°))=(2cos210°)/(4sin10° cos10°)− 2sin10° ((cos25° −sin25°)/(2sin5°cos5°))
=(cos10°)/(2sin10°)−2cos10° ＝(cos10° −2sin20°)/(2sin10°)
=(cos10°−2sin(30°−10°))/(2sin10°)＝(cos10°−2sin30°cos10° +2cos30° sin10°)/(2sin10°)
=cos30° ＝(√(3))/(2)